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The modulus of a complex number a + bi is the square root of a squared plus b squared. Here a = 3 and b = -4, so |z| equals the square root of 9 plus 16, which is the square root of 25, giving 5.
The product rule states that the derivative of u times v is u prime times v plus u times v prime. Taking u = x^3 and v = sin(x) gives u prime = 3x^2 and v prime = cos(x). Substituting produces x^3 cos(x) + 3x^2 sin(x).
An antiderivative of 2x is x squared. Evaluating x squared at the upper limit 3 gives 9, and at the lower limit 0 gives 0. Subtracting gives 9 minus 0, which equals 9.
Acceleration is the derivative of velocity with respect to time. Differentiating 3t squared gives 6t. Substituting t = 2 gives 12, so the acceleration is 12 metres per second squared.
This is a separable first-order equation; dividing by y and integrating gives ln y = 2x + C. Exponentiating both sides yields y = A e^(2x), where A is an arbitrary constant. Thus the general solution is y = C e^(2x).
Expanding the square gives 1 + 2i + i squared. Since i squared equals negative one, this becomes 1 + 2i - 1, which simplifies to 2i.
An infinite geometric series with first term a and common ratio r, where the magnitude of r is less than one, sums to a divided by (1 - r). Here a = 1 and r = 1/2, so the sum is 1 divided by one half, which equals 2.
The determinant of a two-by-two matrix with entries a, b in the first row and c, d in the second is ad minus bc. Here that is 2 times 3 minus 1 times 4, which is 6 minus 4, giving 2.
Stationary points occur where f'(x) = 0, that is 3x(x - 2) = 0, giving x = 0 or x = 2. The second derivative is 6x - 6, which is positive only at x = 2. Hence the required point is x = 2.
The point lies in the second quadrant, since the real part is negative and the imaginary part positive. The reference angle is arctan(1/1) = pi/4, and in the second quadrant the argument is pi minus pi/4, which equals 3pi/4.
With u = x squared plus 1, du = 2x dx, so the integral becomes the integral of u cubed from u = 1 to u = 2. An antiderivative is u to the fourth over 4, giving (16 - 1)/4, which equals 15/4, not a whole number, so rechecking: the antiderivative evaluated gives 15/4. The correct value is therefore 15/4.
Velocity is the derivative of displacement, giving v(t) = 3t^2 - 12t + 9, which factors as 3(t - 1)(t - 3). Setting velocity to zero gives t = 1 or t = 3, and the first time the particle comes to rest is t = 1.
Applying the quadratic formula gives z = (4 plus or minus the square root of (16 - 52)) / 2. The discriminant is -36, whose square root is 6i, so z = (4 plus or minus 6i)/2, which simplifies to 2 plus or minus 3i.
For parametric equations, dy/dx equals (dy/dt) divided by (dx/dt). Here dy/dt = 3t^2 and dx/dt = 2t, so dy/dx = 3t^2/(2t), which simplifies to 3t/2 for t not equal to zero.
The general term is the binomial coefficient of five choose k times (2x)^k. For the x squared term, k = 2, giving ten times four x squared, which is 40 x squared. The coefficient is therefore 40.
The inverse of a two-by-two matrix is the adjugate divided by the determinant. The determinant is 3 times 4 minus 1 times 2, which is 10, and the adjugate swaps the diagonal entries and negates the off-diagonal ones, giving [[4, -1], [-2, 3]]. Dividing by 10 gives the inverse.
Separating variables gives y dy = x dx, and integrating gives y squared over two equals x squared over two plus a constant. Multiplying by two and using y = 3 at x = 0 gives the constant as nine, so y squared = x squared + 9. Taking the positive root gives y = sqrt(x^2 + 9).
The horizontal component equals the magnitude multiplied by the cosine of the angle to the horizontal. Here that is 10 times cos(60 degrees), and cos(60 degrees) equals 0.5. The horizontal component is therefore 5 newtons.
The sequence is arithmetic with first term 2 and twentieth term 59. The sum of an arithmetic sequence is the number of terms times the average of the first and last terms, giving 20 times (2 + 59)/2. This equals 20 times 30.5, which is 610, so the sum is 610.
The absolute value function is non-negative, so its minimum value is zero, occurring where the expression inside is zero. Setting x - 2 = 0 gives x = 2, and the corresponding y value is 3. Thus the graph has a minimum at the point (2, 3).
The velocity vector is the derivative of the position vector, giving (2t, 3). At t = 2 this is (4, 3). The speed is the magnitude of the velocity, which is the square root of (16 + 9), equal to the square root of 25, namely 5.
The chain rule gives the derivative of ln(u) as u prime divided by u. Here u = cos x, so u prime = -sin x, and the derivative is -sin x divided by cos x. This simplifies to -tan x.
Differentiating implicitly gives 2x + y + x(dy/dx) + 2y(dy/dx) = 0. Substituting x = 1 and y = 2 gives 2 + 2 + (dy/dx) + 4(dy/dx) = 0, so 4 + 5(dy/dx) = 0. Hence dy/dx equals negative four fifths.
De Moivre's theorem states that raising cos x plus i sin x to a power n gives cos(nx) plus i sin(nx). Applying this with n = 4 gives cos 4x plus i sin 4x directly.
The volume of revolution about the x-axis is the integral of pi times y squared with respect to x. Since y squared equals x, the integral from 0 to 4 of pi x is pi times x squared over two evaluated at 4. This gives pi times 8, which equals 8pi.
The modulus is |z| = √(4 + 4) = 2√2, and the argument satisfies tan θ = 2/2 = 1 with z in the first quadrant, so θ = π/4. Polar form is r(cos θ + i sin θ), giving 2√2(cos(π/4) + i sin(π/4)).
Apply the recurrence step by step: a(2) = 2(1) + 1 = 3, then a(3) = 2(3) + 1 = 7, then a(4) = 2(7) + 1 = 15. Each term depends on the immediately preceding term, so all intermediate values must be computed.
The dot product of (a, b) and (c, d) equals ac + bd, so here it is 3×4 + (-2)×5 = 12 - 10 = 2. The signs matter because the second component of the first vector is negative.
Setting f'(x) = 3x^2 - 3 = 0 gives x = ±1. The second derivative f''(x) = 6x is negative at x = -1, confirming a local maximum there, while at x = 1 it is positive, indicating a local minimum.
An antiderivative of 3x^2 is x^3, so the integral equals [x^3] evaluated from 1 to 3, which is 27 - 1 = 26. The fundamental theorem of calculus requires subtracting the value at the lower limit.
Conditioning on an even result restricts the sample space to {2, 4, 6}, which has three equally likely outcomes. Of these, 4 and 6 are greater than 2, so the conditional probability is 2/3.
The modulus of a + bi is √(a^2 + b^2), so here it is √(9 + 16) = √25 = 5. The modulus is always a non-negative real number.
For a linear equation dy/dx + P(x)y = Q(x), the integrating factor is exp(∫P dx). Here P = 2/x, so ∫P dx = 2 ln x = ln(x^2), and exponentiating gives x^2.
The determinant of M is (1)(4) - (2)(2) = 0, and a square matrix is singular exactly when its determinant is zero. Geometrically the columns are parallel, so M collapses the plane onto a line and cannot be inverted.
With u = sin x, du = cos x dx, so the integral becomes ∫u^2 du = u^3/3 + C. Substituting back gives (1/3) sin^3 x + C, which differentiates to sin^2 x cos x as required.
In polar form z = √2 (cos(π/4) + i sin(π/4)), so by De Moivre z^6 = (√2)^6 (cos(6π/4) + i sin(6π/4)). Now (√2)^6 = 8 and 6π/4 = 3π/2, giving 8(0 - i) = -8i.
From the equations, cos θ = x/2 and sin θ = y/3. Using the identity cos^2 θ + sin^2 θ = 1 gives x^2/4 + y^2/9 = 1, an ellipse.
The vertical velocity is 15 m/s, and maximum height occurs when vertical velocity is zero: h = v^2/(2g) = 225/(2×9.8) ≈ 11.5 m. The horizontal component does not affect the height.
Differentiating velocity v(t) = (4t, 3t^2) gives acceleration a(t) = (4, 6t). At t = 1 this is (4, 6), whose magnitude is √(16 + 36) = √52 = 2√13.
The general term is C(6, k) 2^(6-k) (-x)^k. For x^3 we take k = 3, giving C(6,3) × 2^3 × (-1)^3 = 20 × 8 × (-1) = -160. The alternating sign comes from the negative term in the binomial.
In polar form 8 has modulus 2 and argument 0, so the cube roots have modulus 2 and arguments 0, 2π/3, 4π/3. These give 2, 2(cos 2π/3 + i sin 2π/3) = -1 + √3 i, and its conjugate -1 - √3 i.
For rotation about the x-axis the disc method gives V = π ∫ y^2 dx over the stated limits. The function must be squared inside the integral and multiplied by π outside.
Eigenvalues satisfy det(A - λI) = 0, which gives (4 - λ)(3 - λ) - (1)(2) = 0. Expanding yields λ^2 - 7λ + 10 = 0, so the eigenvalues are 5 and 2.
Resolving: the first force gives (20cos30, 20sin30) = (17.32, 10) and the second gives (0, 10). The resultant is (17.32, 20), with magnitude √(300 + 400) ≈ 29.1 N.
At (1, 1) the gradient is 1 - 1 = 0. Euler's method gives y(1.5) ≈ y(1) + h × gradient = 1 + 0.5 × 0 = 1.0. The flat slope means the estimate does not change over the step.
Factoring gives (x - 1)(x + 1)/(x - 1). For x ≠ 1 this equals x + 1, so the graph is a line with a single missing point at x = 1, where the limit is 2. A common factor cancels, so no vertical asymptote occurs.
The velocity is zero at t = 0 and t = 6 and is positive in between, so distance equals the displacement. The integral of 6t - t^2 is 3t^2 - t^3/3, evaluated from 0 to 6 giving 108 - 72 = 36 m.
Since z has modulus 1, 1/z equals the conjugate cos θ - i sin θ. Adding gives 2 cos θ because the imaginary parts cancel, a standard result linking complex numbers to trigonometric identities.
Implicit differentiation gives 2x + 2y dy/dx = 0, so dy/dx = -x/y = -3/4 at (3, 4). The normal is perpendicular to the tangent, so its gradient is the negative reciprocal 4/3, giving y - 4 = (4/3)(x - 3)... which matches the reciprocal relationship.
The matrix sends (1, 0) to (0, 1) and (0, 1) to (-1, 0), which is exactly a 90 degree anticlockwise rotation. Its determinant is 1 and its columns are orthonormal, confirming a rotation rather than a reflection or scaling.